`int_(0)^(4) {|x-1| + |x-3| dx}`=

Options

A. 5
B. 8
C. 10
D. none

Answer

To solve the integral \( \int_{0}^{4} \left( |x-1| + |x-3| \right) \, dx \), we will break it down into parts based on the points where the absolute values change, which are \( x = 1 \) and \( x = 3 \).

### Step 1: Identify intervals based on absolute values
The absolute value function changes at the points \( x = 1 \) and \( x = 3 \). Therefore, we will break the integral into three parts:
1. From \( 0 \) to \( 1 \)
2. From \( 1 \) to \( 3 \)
3. From \( 3 \) to \( 4 \)

### Step 2: Rewrite the integral in each interval
1. For \( 0 \leq x < 1 \):
- \( |x - 1| = 1 - x \)
- \( |x - 3| = 3 - x \)
- Thus, \( |x - 1| + |x - 3| = (1 - x) + (3 - x) = 4 - 2x \)

2. For \( 1 \leq x < 3 \):
- \( |x - 1| = x - 1 \)
- \( |x - 3| = 3 - x \)
- Thus, \( |x - 1| + |x - 3| = (x - 1) + (3 - x) = 2 \)

3. For \( 3 \leq x \leq 4 \):
- \( |x - 1| = x - 1 \)
- \( |x - 3| = x - 3 \)
- Thus, \( |x - 1| + |x - 3| = (x - 1) + (x - 3) = 2x - 4 \)

### Step 3: Set up the integral
Now we can set up the integral as follows:
\[ \int_{0}^{4} \left( |x-1| + |x-3| \right) \, dx = \int_{0}^{1} (4 - 2x) \, dx + \int_{1}^{3} 2 \, dx + \int_{3}^{4} (2x - 4) \, dx \]

### Step 4: Calculate each integral
1. **First integral**:
\[ \int_{0}^{1} (4 - 2x) \, dx = \left[ 4x - x^2 \right]_{0}^{1} = (4 \cdot 1 - 1^2) - (4 \cdot 0 - 0^2) = 4 - 1 = 3 \]

2. **Second integral**:
\[ \int_{1}^{3} 2 \, dx = \left[ 2x \right]_{1}^{3} = 2 \cdot 3 - 2 \cdot 1 = 6 - 2 = 4 \]

3. **Third integral**:
\[ \int_{3}^{4} (2x - 4) \, dx = \left[ x^2 - 4x \right]_{3}^{4} = (4^2 - 4 \cdot 4) - (3^2 - 4 \cdot 3) = (16 - 16) - (9 - 12) = 0 - (-3) = 3 \]

### Step 5: Combine the results
Now, we combine the results of the three integrals:
\[ \int_{0}^{4} \left( |x-1| + |x-3| \right) \, dx = 3 + 4 + 3 = 10 \]

### Final Answer
Thus, the value of the integral is \( \boxed{10} \).